Final answer:
The number of different arrangements for 5 boys and 3 girls to be seated in a row, with boys at both ends and no girls together, is 720.
Step-by-step explanation:
The subject question asks about the number of different arrangements possible for seating 5 boys and 3 girls in a row with specific conditions: the leftmost and rightmost seats must be occupied by boys, and no two girls can sit next to each other. First, we place the boys on the positions, with two fixed positions at the ends of the row and three boys to be arranged in between. Therefore, we can arrange the 3 remaining boys in the 3 remaining spaces, which is a permutation problem of 3 boys taken 3 at a time (3!).
Since no girls can sit next to each other, they must sit in the spaces between the boys. There are four spaces (to the left of each of the three boys, plus one space to the right of the third boy in the sequence) where we can place the three girls. Thus, the number of ways to arrange the girls is the permutation of 4 spaces taken 3 at a time (P(4,3)).
Therefore, the total number of arrangements would be the product of both permutations: 5 × 3! × P(4,3). Simplifying, we find that there are 5 × 6 × 24 = 720 different seating arrangements possible under the given conditions.